Using multiple conditions within an if statement

Question

How using multiple conditions within an if statement?

 function testNum(a) { if (a == (1 || 2 || 3)) { return "is 1, 2, or 3"; } else { return "is not 1, 2, or 3"; } } console.log(testNum(1)); // returns "is 1, 2, or 3" console.log(testNum(2)); // returns "is not 1, 2, or 3" console.log(testNum(3)); // returns "is not 1, 2, or 3" 

testNum(2) and testNum(3) should return: "is 1, 2 or 3" but doesn't.

Answer 1

In this particular scenario, you can even use an array and Array#includes method for checking.

if ([1, 2, 3].includes(a)) {
  // your code
}

 function testNum(a) { if ([1, 2, 3].includes(a)) { return "is 1, 2, or 3"; } else { return "is not 1, 2, or 3"; } } console.log(testNum(1)); console.log(testNum(2)); console.log(testNum(4)); console.log(testNum(3)); 

---

FYI : In your current code (1 || 2 || 3) results 1 (since 1 is truthy) and actually a == (1 || 2 || 3) does a == 1 . The right way is to seperate each conditions with || (or), for eg : a == 1 || a == 2 || a ==3 a == 1 || a == 2 || a ==3 a == 1 || a == 2 || a ==3 .

For more details visit MDN documentation of Logical operators .

Answer 2

You cannot have || like that. The one you have used is not the right way. You should be using:

 function testNum(a) { if (a == 1 || a == 2 || a == 3) { return "is 1, 2, or 3"; } else { return "is not 1, 2, or 3"; } } console.log(testNum(1)); console.log(testNum(2)); console.log(testNum(3)); 

Answer 3

Your or-operators are placed incorrectly:

function testNum(a) {
    if (a == 1 || a == 2 || a == 3) {
        return "is 1, 2, or 3";
    } else {
        return "is not 1, 2, or 3";
    }
}

Before, you were testing if a was equal to 1 || 2 || 3 1 || 2 || 3 1 || 2 || 3 which evaluates to 1 †. So you were just checking a == 1 which is not what you wanted!

† Essentially when you string together "or"s like this, the first truthy value is returned. For example, you can assert for yourself that: 0 || False || 5 0 || False || 5 0 || False || 5 gives 5 .

Answer 4

You can try using a list of qualified values and check if the lookup value belongs there:

function testNum(a) {
    var candidates = [1, 2, 3];
    if (candidates.indexOf(a) != -1) {
        return "is 1, 2, or 3";
    } else {
        return "is not 1, 2, or 3";
    }
}

console.log(testNum(1)); // returns "is 1, 2, or 3"
console.log(testNum(2)); // returns "is not 1, 2, or 3"
console.log(testNum(3));
console.log(testNum(4));

Answer 5

You can either

    if (a == 1 || a == 2 || a == 3) {

or

    if ([1, 2, 3].includes(a)) {

which will be more convenient when there's more values to test. If you want to be compatible with ancient browsers, you have to write it like this

    if ([1, 2, 3].indexOf(a) >= 0) {

which look undeniably uglier.

Answer 6

The reason why you are not getting the expected result because you are performing a Logical OR operation here.

As per MDN Network : Logical OR (||) expr1 || expr2 If expr1 can be converted to true, returns expr1; else, returns expr2.

(1 || 2 || 3) is evaluated and the operation results at 1. You can check the result by pasting (1 || 2 || 3) in Chrome debugger or something.

So when the first number is 1 in all cases, the expression result is also 1. You can change the expression as how others have advised to get the desired output.